# Ordinary diff. eq

## Main Question or Discussion Point

I wish to transform my diff. eq.
(1/r)*(dy/dr)*(d2/dr2(r*y))*dr into a more convenient expression, in a similar to the following transformation:

(dy/dx)*(d2y/dx2)*dx = 0.5*(d/dx(dy/dx)^2)*dx
which is a very convenient expression for integration ==> 0.5* (dy/dx)^2

So far, I have found the following expression, to which I haven't found the integration answer.
(1/(2*x^4)*(d/dx(x^2*dy/dx)^2)

tbk1

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mfb
Mentor
If you want to solve something, you need an equation. I do not see any "=" in your problem.
Your first expression has an additional dr which I would not expect there.

Converted to LaTeX, as it is easier to read:
$$\frac{1}{r} \left(\frac{dy}{dr}\right) \left(\frac{d^2}{dr^2} (ry)\right) dr$$
$$\frac{1}{2x^4}\frac{d}{dx} \left(x^2 \frac{dy}{dx}\right)^2$$

Thank you for your quick reply. I do not LaTex, so till I'll find a quick way to translate it, I wish to make the following remarks:
1. You're right, a differential, dr, is missing of course.
2. You may address it as an homogeneous equation, but the truth is that I simply seek the simpler expression to integrate, something like d(f(x))dx, so that the integral result will be
3. I mixed "x" and "r", all the independent variables should be either "x" or "r".
Thanks again
tbk1

mfb
Mentor
The second equation looks good in that respect. Multiply with x^4, and you can integrate both sides. The left side is trivial (as it is df(x)/dx), the right side is 0 or something you can integrate, and afterwards you can try to simplify the result to integrate again.

oops, reading your reply, I understand that one cannot address the equation as a homogeneous one, but rather as equal to a constant, than the term "x^4" cannot be eliminated.

I wish to transform my diff. eq.
(1/r)*(dy/dr)*(d2/dr2(r*y))*dr into a more convenient expression, in a similar to the following transformation:

(dy/dx)*(d2y/dx2)*dx = 0.5*(d/dx(dy/dx)^2)*dx
which is a very convenient expression for integration ==> 0.5* (dy/dx)^2

So far, I have found the following expression, to which I haven't found the integration answer.
(1/(2*x^4)*(d/dx(x^2*dy/dx)^2)

tbk1
Hi !
Fishy wording ! I cannot understand exactly what is the equation.
Would you mind rewrite only the first equation, without explanation nor comment which could confused us. Only one equation on the patern :
(1/r)*(dy/dr)*(d2/dr2(r*y)) = what ?

mfb
Mentor
oops, reading your reply, I understand that one cannot address the equation as a homogeneous one, but rather as equal to a constant, than the term "x^4" cannot be eliminated.
It cannot be eliminated, but it is easy to integrate it with respect to x.

Hi,
I have a question regarding to the series solution of ordinary differential equation . my question is , I've found that if we have to solve y"+p(x)y'+q(x)y=0 by power series method, we have to find a point say "a" where p(x) & q(x) become analytic , and by the definition of analytic function a function will be analytic if it has a taylor series expansion around a given neighbourhood point and for taylor series expansion the function must be infinitely differentiable but for the equation y"+xy'+x^2y=0. how x & x^2 become analytic as by the 2 differentiation of x and 3 differentiation of x^2 it becomes zero , and so there not infinitely differentiable ?

how x & x^2 become analytic as by the 2 differentiation of x and 3 differentiation of x^2 it becomes zero , and so there not infinitely differentiable ?
The first derivative of x² is 2x
The second derivative is 2
The third derivative is 0 because the derivative of any constant function is 0.
The fourth derivative is 0 because the derivative of any constant function is 0.
And so on. All successive derivatives are 0.
So, x² is infinitely differentiable.

The first derivative of x² is 2x
The second derivative is 2
The third derivative is 0 because the derivative of any constant function is 0.
The fourth derivative is 0 because the derivative of any constant function is 0.
And so on. All successive derivatives are 0.
So, x² is infinitely differentiable.
Thanx